I don't know how to find the normal force on this problem

In summary, the problem involves a 3.0 kg wood box sliding down a vertical wood wall with a force being applied at a 49° angle. To determine the magnitude of force needed to maintain a constant speed, you must use Newton's second law for net force in the x and y directions. The normal force and kinetic friction must be taken into account, with the coefficient of kinetic friction for wood being 0.2. Setting up the appropriate equations and solving for the unknown forces will yield the desired force to maintain constant speed.
  • #1
xstetsonx
78
0
1. A m = 3.0 kg wood box slides down a vertical wood wall while you push on it at a θ = 49° angle. What magnitude of force should you apply to cause the box to slide down at a constant speed?

2. Fk-Mk*n


3. i know there are 4 forces acting on the object. gravity, normal force from the wall, kinetic friction, and the force that i applied. but i do not know how to set up the equation to solve this problem
 
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  • #2
Do you know how to draw a free body diagram?
 
  • #3
yea i did one but the n is on the x-axis so that means i need to know the F to get the normal force right?
 
  • #4
First off, you have to neglect friction since you do not know the kinetic friction coefficient of the surface. Second, the normal force doesn't cause the block to move in what you defined to be the x direction. Are you using Newton's second law for net force in the x and y directions?
 
  • #5
Vykan12 said:
First off, you have to neglect friction since you do not know the kinetic friction coefficient of the surface. Second, the normal force doesn't cause the block to move in what you defined to be the x direction. Are you using Newton's second law for net force in the x and y directions?

o i am sorry this question some how expected us to find the coefficient of kinetic friction on wood which is 0.2. and yes
 
  • #6
Welcome to PF!

Hi xstetsonx! Welcome to PF! :wink:
xstetsonx said:
i need to know the F to get the normal force right?

No, you don't need to know F, you can just call it F.

For constant speed, you need zero acceleration, which means zero net force in both the x and y direction.

So call the force F, find N, and then find µN. :smile:
 
  • #7
so the total force will be
mK*Fcos49-mg+Fsin49?
 
  • #8
xstetsonx said:
so the total force will be
mK*Fcos49-mg+Fsin49?

Yes, except I'm not sure about that m at the beginning.
 
  • #9
You should get

[tex]F_{app} - F_{k} - mg\sin(\theta) = 0[/tex]
[tex]n - mg\cos(\theta) = 0[/tex]

The algebra's up to you.

Edit:

so the total force will be
mK*Fcos48-mg+Fsin48?

Yes, that's correct, assuming F is the force of gravity.
 

1. What is the normal force?

The normal force is the force exerted by a surface on an object that is in contact with it. It is always perpendicular to the surface and acts in the opposite direction of the force applied by the object on the surface.

2. How do I calculate the normal force?

To calculate the normal force, you need to know the weight of the object and the angle at which it is in contact with the surface. The normal force is equal to the weight of the object multiplied by the cosine of the angle.

3. Can the normal force be greater than the weight of the object?

No, the normal force can never be greater than the weight of the object. It can only be equal to or less than the weight of the object, depending on the angle at which the object is in contact with the surface.

4. What happens if there is no normal force?

If there is no normal force, it means that the object is not in contact with a surface. In other words, it is either floating in space or falling freely under the influence of gravity.

5. How does the normal force affect the motion of an object?

The normal force can affect the motion of an object by changing its direction or magnitude. If the normal force is greater than the weight of the object, it can cause the object to accelerate upwards. If the normal force is less than the weight of the object, it can cause the object to accelerate downwards.

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